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I have thought up a new redux variant that skips a whole step, I'm not 100% sure that it works but my idea is this

First, you solve the centres as normal

Second, you solve all the edges except the ones with a yellow sticker on them

Thirdly, you solve it like a regular 3x3, without pll, pretending that the yellow edges are solved (ask if you don't understand this, I might have to explain in more detail)

Lastly, you do one algorithm to permute all the yellow edges

Reply if you would like me to make the algorithms for the lask step
Also comment if you have any questions/need a step explained in more detail...

I really hope this works because I think it will be alot faster than regular redux and it might really have potential


Also I'd like to apologise if this has already been suggested, I looked around the forums and couldn't find it
http://snk.digibase.ca/k4/7.htm

3-look version for the last layer edges. That is already in use in K4 method :)

And yeah I don't understand what is the step that you say you are skipping... you just do last 4 edges at different time than in Yau/redux...
 
New 4x4 Method Tell Me What You Think

Hello I'm a 16 second average on 3x3 and 1:20 average on 4x4 (I don't practice much) this is my first post, sorry if this method has already been made, but I'm really sure it hasn't.
This method has a lot in common with Yau and reduction method but also has big differences in certain parts of the solve. I'm not very good at describing so I will do my best. PAY ATTENTION TO STEPS 3 AND 3 1/2 AND 4! I will be making a video if requested.

1. Build two centers across from each other

2. Line up each center so they are in 1x2 blocks long ways

3. Place edge pieces in the middle layer so one is in the u layer and another is in the d layer, pair them and place it in the bottom layer. At the end of this step you should have four edges that are solved in the bottom layer while making sure the centers stay in 1x2 bars long ways in the u and d layers after each edge

3 1/2. Solve four more edges and place them in the top layer. Make SURE you have all four edges solved that go to the color of your cross. When you tuck away the solved pieces in the top layer make sure they are oriented if they are a cross piece

4. At this point you should have Eight edges solved, four on top and four in the bottom. For this step turn the cube to the two solved centers are on the left and right. for this step you are able to turn l, r, U2, F2, B2, D2. Now solve the centers (This step usually takes me two seconds) ALSO if you made the cross pieces oriented earlier then it will result in a very easy cross after this step

5. For edges will be left, solve them as you please

6. 3x3 phase

7. (PROFIT) I mean you're done.
 
Hello I'm a 16 second average on 3x3 and 1:20 average on 4x4 (I don't practice much) this is my first post, sorry if this method has already been made, but I'm really sure it hasn't.
This method has a lot in common with Yau and reduction method but also has big differences in certain parts of the solve. I'm not very good at describing so I will do my best. PAY ATTENTION TO STEPS 3 AND 3 1/2 AND 4! I will be making a video if requested.

1. Build two centers across from each other

2. Line up each center so they are in 1x2 blocks long ways

3. Place edge pieces in the middle layer so one is in the u layer and another is in the d layer, pair them and place it in the bottom layer. At the end of this step you should have four edges that are solved in the bottom layer while making sure the centers stay in 1x2 bars long ways in the u and d layers after each edge

3 1/2. Solve four more edges and place them in the top layer. Make SURE you have all four edges solved that go to the color of your cross. When you tuck away the solved pieces in the top layer make sure they are oriented if they are a cross piece

4. At this point you should have Eight edges solved, four on top and four in the bottom. For this step turn the cube to the two solved centers are on the left and right. for this step you are able to turn l, r, U2, F2, B2, D2. Now solve the centers (This step usually takes me two seconds) ALSO if you made the cross pieces oriented earlier then it will result in a very easy cross after this step

5. For edges will be left, solve them as you please

6. 3x3 phase

7. (PROFIT) I mean you're done.

EDIT: Misunderstood what that second step was (though still not 100% sure I believe it'll take 2 seconds), but yeah, what TDM said. Already discussed several times.
 
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Pros: Centers are easier than Yau, Don't have look at the bottom layer for pieces after first four edges, Cross very easy cross, No pause between, Centers are also fast to build the 1x2 blocks.

I will post a video in a few hours to better describe the method.
 
Do you mean U and D or u and d?

Probably U and D if referring to centres.

This exists, it's called freeslice.

Correct me if I'm misinformed, but isn't freeslice when you solve the edges around four solved centres and then fix the centres afterwards?

Redux - F2C, L4C, F8E, L4E, 3x3
This method - F2C, F8E, L4C, L4E, 3x3

This is a weird variation of Redux, with a slight Yau-ey look to the L4C.

Maybe this could be edited so when solving the F4E, you solve the white cross? (this would mean it is more of a Yau submethod than a Redux one)

As most very fast cubers (sub-40) use Yau, I think the bottom method will have a better chance than the Redux variant.
 
Really there isn't much difference, and in my opinion there are more disadvantages than advantages to this that just doing freeslice.

What he said. Robert Yau did move count on IRC earlier and found that trying to solve the centers into 1x2 blocks really wasn't saving any moves over just solving the centers. Basically, you're wasting moves for no advantage.
 
What he said. Robert Yau did move count on IRC earlier and found that trying to solve the centers into 1x2 blocks really wasn't saving any moves over just solving the centers. Basically, you're wasting moves for no advantage.

Hmm, are you sure he was being really efficient? Most of the time one trigger (r U r' or similar) can form 2 bars, so I would be surprised if it wasn't at least movecount efficient.
 
I dont know if I put this already. I couldnt find it so...
Anyway this is for solving the 3x3 cube.
1.EO (Edge orientation)
2. F2l-1 ( The missing pair must be in FR or BL)
3. COLL +1 OR solve the corner of the first layer then COLL.
4. Solve 2 adjacent sides on the top layer.
5. Cycle edges to solve the cube. (3 edges left)
 
Hmm, are you sure he was being really efficient? Most of the time one trigger (r U r' or similar) can form 2 bars, so I would be surprised if it wasn't at least movecount efficient.

Based on the move counts he was posting, yes, I am sure he was being really efficient.

I dont know if I put this already. I couldnt find it so...
Anyway this is for solving the 3x3 cube.
1.EO (Edge orientation)
2. F2l-1 ( The missing pair must be in FR or BL)
3. COLL +1 OR solve the corner of the first layer then COLL.
4. Solve 2 adjacent sides on the top layer.
5. Cycle edges to solve the cube. (3 edges left)

This is a ZZ LS-LL subset. I believe this has been suggested before, sort of a reverse Heise LL with a lot more algs.

EDIT: If you're willing to learn COLL+1, you might as well learn PL5E (with EPLL subtracted out, it's only 48 algs, which could be reduced with inverses).
 
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